One common math puzzle I've seen around asks for how many zeros are in the product of "100!"

Usually, the solution everyone gives goes something like try to match pairs of 5s and 2s that factor out of the numbers, which ends up being 24 zeroes (you can factor a 5 out of 20 of the numbers, and factor 2 5s out of 4 of the numbers; you can factor more than 24 2s out).

This however as far as I know gives the number of trailing zeroes at the end of the number, but does not account for the zeroes that are within the number. My question is, is this answer correct anyways? Can there be zeroes that aren't trailing that are inside? Why or why not and if there can be can we somehow figure out how many are within the product?


  • 17
    $\begingroup$ I am not aware of any way to predict the number of non-trailing zeros in $n!$, other than by calculating n! and counting them. $\endgroup$ – Robert Israel May 7 '12 at 7:41
  • 1
    $\begingroup$ See also oeis.org/A137581 $\endgroup$ – Robert Israel May 7 '12 at 7:47
  • 1
    $\begingroup$ According to WolframAlpha it would be $29$ zeros in $100!$ (trailing $24$ and $5$ zeroes inside), but if you are looking for a method, as Robert Israel said, there is no known method. $\endgroup$ – Kirthi Raman May 7 '12 at 11:22
  • 4
    $\begingroup$ I would expect that for large $n$ about one-tenth of the digits should be zero, simply because there's no good reason to expect otherwise. The number of digits of $n!$ can be estimated very well by Stirling's formula. $\endgroup$ – Gerry Myerson Jul 7 '12 at 5:26
  • 1
    $\begingroup$ @robjohn, 100 is not large. $\endgroup$ – Gerry Myerson Jul 7 '12 at 11:42

For a prime $p$, let $\sigma_p(n)$ be the sum of the digits of $n$ when written in base-$p$ form. Then the number of factors of $p$ that divide $n!$ is $$ \frac{n-\sigma_p(n)}{p-1} $$

There are $24$ trailing zeroes in $100!$. Since $100_{\text{ten}}=400_{\text{five}}$, there are $\frac{100-4}{5-1}=24$ factors of $5$ in $100!$.

However, there are $6$ other zeros that occur earlier, making the total $30$:

$933262154439441526816992388562667\color{#C00000}{00}49\color{#C00000}{0}71596826438162146859296389$ $52175999932299156\color{#C00000}{0}894146397615651828625369792\color{#C00000}{0}82722375825118521\color{#C00000}{0}$ $916864\color{#C00000}{000000000000000000000000}$

  • 12
    $\begingroup$ Counting the amount of zeroes from the explicit form kind of kills the point of doing your first observation. $\endgroup$ – Listing Jul 7 '12 at 11:50
  • $\begingroup$ @Listing: It does, but I had written the first part before I pulled out Mathematica to compute $100!$. I left it in because I like the formula, and because I used it in a comment to lab bhattacharjee's answer. $\endgroup$ – robjohn Jul 7 '12 at 14:01
  • 1
    $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Jul 25 '14 at 10:51
  • $\begingroup$ It is funny that zeroes are relatively rare before the final zero part. $\endgroup$ – VividD Apr 19 '15 at 9:21
  • $\begingroup$ +1 from me. You are 185k rep, you know by now, the haters hate. And rarely explain their DV. This was a brilliant answer. If I run into this problem in the future, it will be a pleasure to present this solution. $\endgroup$ – JoeTaxpayer Feb 22 '16 at 14:12

If you want to learn how many trailing zeroes are there in a factorial, it is

$\frac{N}{5} + \frac{N}{5}^2 + \frac{N}{5}^3 ..... \frac{N}{5}^{(m-1)} WHERE (\frac{N}{5}^m)<1$

You can learn here how this formula comes at



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.